Improper Integrals: Type II and Related Concepts

Warm-up: Evaluating an Indefinite Integral with Partial Fractions and Limits

• Original Integral: $\int \frac{1}{(x-3)(x+1)}dx$

• Method: Partial Fraction Decomposition

  • Set up the decomposition: $\frac{1}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1}$

  • Find common denominator: $1 = A(x+1) + B(x-3)$

  • Solve for $A$ and $B$:

    • Set $x=3$: $1 = A(3+1) + B(3-3) \implies 1 = 4A \implies A = \frac{1}{4}$

    • Set $x=-1$: $1 = A(-1+1) + B(-1-3) \implies 1 = -4B \implies B = -\frac{1}{4}$

• Rewrite the Integral: $\int \left( \frac{1}{4(x-3)} - \frac{1}{4(x+1)} \right) dx$

• Integrate:

  • $= \frac{1}{4} \ln|x-3| - \frac{1}{4} \ln|x+1| + C$

  • $= \frac{1}{4} \ln \left| \frac{x-3}{x+1} \right| + C$

• Example Application to an Improper Integral (Implicit): The transcript hints at evaluating a definite integral that is improper, specifically where the upper bound approaches $3$. If we consider $\int_{0}^{3} \frac{1}{(x-3)(x+1)}dx$, it would be defined as:

  • $\lim<em>{t \to 3^-} \int</em>{0}^{t} \frac{1}{(x-3)(x+1)}dx$

  • Evaluating the definite integral from $0$ to $t$:

    • $\left[ \frac{1}{4} \ln \left| \frac{x-3}{x+1} \right| \right]_{0}^{t} = \frac{1}{4} \ln \left| \frac{t-3}{t+1} \right| - \frac{1}{4} \ln \left| \frac{0-3}{0+1} \right|$

    • $= \frac{1}{4} \ln \left| \frac{t-3}{t+1} \right| - \frac{1}{4} \ln 3$

  • Evaluate the limit as $t \to 3^-$:

    • As $t \to 3^-$, $(t-3)$ approaches a small negative number from the left (i.e., $(3-\epsilon)-3 = -\epsilon$). But since it is inside an absolute value, it approaches a small positive number. $(t+1)$ approaches $4$.

    • Therefore, $\lim_{t \to 3^-} \frac{1}{4} \ln \left| \frac{t-3}{t+1} \right| = \frac{1}{4} \ln \left( \frac{\text{small positive number}}{4} \right) = \frac{1}{4} ( -\infty) = -\infty$

  • Conclusion: Since the limit is $-\infty$, this improper integral diverges.

• L'Hôpital's Rule and Indeterminate Forms: The warm-up also briefly touches on indeterminate forms relevant to improper integrals.

  • An expression like $\lim_{t \to 0^+} t \ln(t)$ is an indeterminate form of type $0 \cdot (- \infty)$.

  • To apply L'Hôpital's Rule, it must be rewritten as a fraction: $\lim_{t \to 0^+} \frac{\ln(t)}{1/t}$, which is of type $\frac{-\infty}{\infty}$.

  • Applying L'Hôpital's Rule: $\lim<em>{t \to 0^+} \frac{1/t}{-1/t^2} = \lim</em>{t \to 0^+} (-t) = 0$

Type II Improper Integrals: Infinite Discontinuity

• Definition: Type II improper integrals occur when the integrand $f(x)$ has an infinite discontinuity at one or both of the limits of integration ($a$ or $b$) or at a point $c$ within the interval $[a,b]$.

  • An "infinite discontinuity" means that the function approaches $\pm \infty$ at that point, making it undefined.

• Case 1: Discontinuity at the Lower Bound ($x=a$)

  • Given $\int_{a}^{b} f(x)dx$, if $f(x)$ is undefined at $x=a$, we rewrite the integral using a limit:

    • $\int<em>{a}^{b} f(x)dx = \lim</em>{t \to a^+} \int_{t}^{b} f(x)dx$

• Case 2: Discontinuity at the Upper Bound ($x=b$)

  • Given $\int_{a}^{b} f(x)dx$, if $f(x)$ is undefined at $x=b$, we rewrite the integral using a limit:

    • $\int<em>{a}^{b} f(x)dx = \lim</em>{t \to b^-} \int_{a}^{t} f(x)dx$

• Case 3: Discontinuity within the Interval ($x=c$ where a < c < b)

  • If $f(x)$ is undefined at an interior point $c$, we split the integral into two parts:

    • $\int<em>{a}^{b} f(x)dx = \int</em>{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx$

    • Each part is then handled as in Case 1 or Case 2: $\lim<em>{t \to c^-} \int</em>{a}^{t} f(x)dx + \lim<em>{u \to c^+} \int</em>{u}^{b} f(x)dx$

    • For the original integral to converge, BOTH resulting limits must converge.

• Solving Procedure (Same as Type I):

  1. Evaluate the definite integral normally, using $t$ as one of the bounds.

  2. Evaluate the limit of the result as $t$ approaches the point of discontinuity.

  3. If the limit exists as a finite number, the integral converges. Otherwise, it diverges.

Example: Evaluating $\int_{0}^{1} \ln xdx$

• Identify Discontinuity: The function $f(x) = \ln x$ is undefined as $x \to 0^+$ (where it approaches $- \infty$). This is a discontinuity at the lower bound $(a=0)$. Therefore, we use Case 1.

• Rewrite with a Limit:

  • $\int<em>{0}^{1} \ln xdx = \lim</em>{t \to 0^+} \int_{t}^{1} \ln xdx$

• Evaluate the Indefinite Integral ($\int \ln xdx$):

  • Use Integration by Parts: $\int u dv = uv - \int v du$

    • Let $u = \ln x \implies du = \frac{1}{x} dx$

    • Let $dv = dx \implies v = x$

  • $\int \ln xdx = (\ln x)(x) - \int (x)\left( \frac{1}{x} \right) dx$

  • $= x \ln x - \int 1 dx$

  • $= x \ln x - x + C$

• Evaluate the Definite Integral from $t$ to $1$:

  • $\left[ x \ln x - x \right]_{t}^{1} = (1 \ln 1 - 1) - (t \ln t - t)$

  • Since $\ln 1 = 0$: $= (0 - 1) - (t \ln t - t)$

  • $= -1 - t \ln t + t$

• Evaluate the Limit:

  • $\lim_{t \to 0^+} (-1 - t \ln t + t)$

  • Separate the terms: $\lim<em>{t \to 0^+} (-1) - \lim</em>{t \to 0^+} (t \ln t) + \lim_{t \to 0^+} (t)$

  • We know that $\lim_{t \to 0^+} (-1) = -1$

  • We know from L'Hôpital's Rule (as demonstrated in the warm-up) that $\lim_{t \to 0^+} (t \ln t) = 0$

  • We know that $\lim_{t \to 0^+} (t) = 0$

  • Substitute these values: $= -1 - 0 + 0 = -1$

• Conclusion: Since the limit exists and is a finite number ($-1$), the improper integral $\int_{0}^{1} \ln xdx$ converges to $-1$.